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3 years ago

Most chemical reactions can be divided into how many main groups?

1 answer:

5 0

There is synthesis
decomposition
double displacement
single displacement
combustion
metathesis
so i guess you could say 6

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1. A student adds water and sugar to a jar and seals the jar so that nothing can get in or out. The

Tcecarenko [31]

Answer:

The mass will stay the same throughout time

3 0

2 years ago

The centripetal force acting on the space shuttle

tamaranim1 [39]

Answer:

(4) weight

Explanation:

The centripetal force acting on the space shuttle in orbit is given by:

$F=m\frac{v^2}{r}$

where

m is the mass of the shuttle

v is the tangential speed of the shuttle

r is the radius of its circular orbit

When the shuttle orbits the Earth, the centripetal force that keeps the shuttle in circular motion is given by the gravitational attraction between the shuttle and the Earth, which corresponds to the weight of the shuttle, and it is given by:

$F=G\frac{Mm}{r^2}$

where

G is the gravitational constant

M is the Earth's mass

And this force, therefore, corresponds to the centripetal force.

7 0

3 years ago

Read 2 more answers

A particle with charge 3.01 µC on the negative x axis and a second particle with charge 6.02 µC on the positive x axis are each

ra1l [238]

Answer:

The third particle should be at 0.0743 m from the origin on the negative x-axis.

Explanation:

Let's assume that the third charge is on the negative x-axis. So we have:

$E_{1}+E_{3}-E_{2}=0$

We know that the electric field is:

$E=k\frac{q}{r^{2}}$

Where:

k is the Coulomb constant
q is the charge
r is the distance from the charge to the point

So, we have:

$k\frac{q_{1}}{r_{1}^{2}}+k\frac{q_{3}}{r_{3}^{2}}-k\frac{q_{2}}{r_{2}^{2}}=0$

Let's solve it for r(3).

$\frac{3.01}{0.0429^{2}}+\frac{9.03}{r_{3}^{2}}-\frac{6.02}{0.0429^{2}}=0$

$r_{3}=0.0743\:$

Therefore, the third particle should be at 0.0743 m from the origin on the negative x-axis.

I hope it helps you!

3 0

3 years ago

Suppose that a planet were discovered between the sun and Mercury, with a circular orbit of radius equal to 2/3 of the average o

Musya8 [376]

Explanation:

It is given that,

A planet were discovered between the sun and Mercury, with a circular orbit of radius equal to 2/3 of the average orbit radius of Mercury.

Mass of the Sun, $M=1.99\times 10^{30}\ kg$

Radius of Mercury's orbit, $r=5.79\times 10^{10}\ m$

Radius of discovered planet, $R=\dfrac{2}{3}r$

$R=\dfrac{2}{3}\times 5.79\times 10^{10}\ m=3.86\times 10^{10}\ m$

Let T is the orbital period of such a planet. Using Kepler's third law of planetary motion as :

$T^2\propto R^3$

$T^2=\dfrac{4\pi^2R^3}{GM}$

$T^2=\dfrac{4\pi^2\times (3.86\times 10^{10})^3}{6.67\times 10^{-11}\times 1.99\times 10^{30}}$

$T=\sqrt{1.71\times 10^{13}}$

T = 4135214.625 s

T = 47.86 days

So, the orbital period of such a planet is 47.86 days. Hence, this is the required solution.

6 0

3 years ago

Why doesn’t the moon fall toward the earth and smash into it?

yarga [219]

Gravity

The moon doesn't smash into the earth because the gravity from the earth keeps the moon in orbit around it.

8 0

3 years ago